All categories

2 years ago

Tammys chocolate bar is 59% cocoa.If the weight of the chocolate bar is 63 grams ,how many grams of cocoa does it contain?

Mathematics

1 answer:

QveST [7]2 years ago

3 0

Answer:

37.17 grams

Step-by-step explanation:

59% of 63 grams = 0.59 × 63 grams = 37.17 grams

You might be interested in

A worker earns $9 per hour and works 25 hours each week. If the worker receives a 3% pay increase, how much additional pay will

Ber [7]

Answer: $6.75

Step-by-step explanation:

25 hours times $9= $225/week

3% increase from $9.00= $0.27

$9.27 times 25 hours= $231.75

$231.75-$225= $6.75 more per week

5 0

3 years ago

Renna pushes the elevator button, but the elevator does not move. The mass limit for the elevator is

earnstyle [38]

Answer:

Whats the full question ?

Step-by-step explanation:

8 0

3 years ago

Simplify (11 + 5i)(11 - 51)

Ede4ka [16]

Answer:

(146 + 0i)

Step-by-step explanation:

(11+5i)(11-5i)

The formula is :

(a+bi) • (x+yi) = ax+by•i^2+(ay+bx)i

and since i^2 = -1 it can be written as :

ax - by + (ay + bx) i

( 11 + 5i) • ( 11 - 5i) =

(146 + 0i)

hope this helped brainlest owo???

7 0

3 years ago

Read 2 more answers

What is the number in scientific notation? 0.000013

leva [86]

13^6 you whould have to move the decimal over 6 times :)

5 0

3 years ago

Read 2 more answers

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

3 0

3 years ago